(1)Calculus/analyse najaar 2007 Uitwerkingen huiswerk week 5 Opgave 17

Calculus/analyse najaar 2007

Uitwerkingen huiswerk week 5

Opgave 17.

Bepaal primitieven F (x) voor de volgende functies:

(i) f (x) := 1

1 + x, (ii) f (x) := x

1 + x, (iii) f (x) := a^x

b^x met a, b > 0, a, b 6= 1, (iv) f (x) := 1

√a²− x², (v) f (x) := 1

√x− 1 +√

x+ 1, (vi) f (x) := 1 1 + sin(x), Oplossing.

(i) F (x) = ln(1 + x);

(ii) f (x) = ^x_1+x⁺¹ −_1+x¹ = 1 −_1+x¹ ⇒ F (x) = x − ln(1 + x);

(iii) f (x) = (^a_b)^x = exp(ln(^a_b)x) ⇒ F (x) = _ln(^1a

b)exp(ln(^a_b)x) = ln(a)−ln(b)¹ (^a_b)^x; (iv) arcsin(x)^′ = ^{√ 1}

1−x² en f (x) = ¹a √ 1

1−(^xa)² ⇒ F (x) = arcsin(^xa);

(v) f (x) = √ ¹ x−1+√x+1

√x+1−√

√x+1−√x−1

x−1 = ^√(x+1)−(x−1)^{x+1−√x−1} = ¹₂(√

x+ 1 −p(x − 1)) ⇒ F(x) = ¹₃((x + 1)³² − (x − 1)³²);

(vi) f (x) = _1+sin(x)^{1 1−sin(x)}_1−sin(x) = ^1−sin(x)

1−sin²(x) = _cos¹2(x) − _cos^sin(x)2_(x). Verder geldt tan(x)^′ = (_cos(x)^sin(x))^′ = _cos¹2

(x)en (_cos(x)¹ )^′ = _cos^sin(x)2

(x) ⇒ F (x) = tan(x)−_cos(x)¹ . Opgave 18.

Bereken de volgende integralen:

(i) Z 1

0 (1 − x)ⁿdxvoor n ∈ N (ii) Z ^π

0 sin(mx) dx voor m ∈ Z.

Oplossing.

(i) f (x) = (1−x)ⁿ ⇒ F (x) = n⁻¹+1(1−x)ⁿ⁺¹⇒R1

0(1−x)ⁿdx= F (1)−F (0) =

1 n+1.

(ii) f (x) = sin(mx) ⇒ F (x) = ⁻¹m cos(mx) ⇒ R^π

0 sin(mx) dx = F (π) − F(0) =

(₂

m als m oneven 0 als m even.

Opgave 19.

Bepaal de volgende integralen door parti¨ele integratie:

(i) Z

x²e^x dx, (ii) Z √

x ln(x) dx, (iii) Z

ln²(x) dx, (iv)

Z

ln³(x) dx, (v) Z

cos(ln(x)) dx, (vi) Z

x arctan(x) dx.

Oplossing.

(i) R x²e^x dx= x²e^x−R 2xe^x dx= x²e^x− 2xe^x+R 2e^x dx= (x²− 2x + 2)e^x; (ii) R √x ln(x) dx = ²₃x³²ln(x) − ²₃R√x dx= ²₃x³²ln(x) − ⁴₉x³²;

(iii) R ln²(x) dx = (ln(x)x − x) ln(x) −R (ln(x) − 1) dx = (ln(x)x − x) ln(x) − (ln(x)x − x) + x = ln²(x)x − 2 ln(x)x + 2x,

(iv) R ln³(x) dx = (ln(x)x − x) ln²(x) −R (ln(x) − 1)2 ln(x) dx = (ln(x)x − x) ln²(x)−2(ln²(x)x−2 ln(x)x+2x)+2(ln(x)x−x) = ln³(x)x−3 ln²(x)x+

6 ln(x)x − 6x;

(v) R cos(ln(x)) dx = x cos(ln(x))+R sin(ln(x)) = x cos(ln(x))+x sin(ln(x))−

R cos(ln(x)) = ¹₂x(cos(ln(x)) + sin(ln(x)));

(vi) R x arctan(x) dx = ¹₂x²arctan(x) −¹₂R x² 1+x² dx

= ¹₂x²arctan(x) − ¹₂R x²+1

1+x² dx+ ¹₂R ₁

1+x² dx = ¹₂x²arctan(x) − ¹₂x+

1

2arctan(x).

Opgave 20.

Bewijs de volgende reductie formules (m.b.v. parti¨ele integratie):

(i) Z

sinⁿ(x) dx = −1

nsinⁿ⁻¹(x) cos(x) +n− 1 n

Z

sinⁿ⁻²(x) dx;

(ii) Z

cosⁿ(x) dx = 1

ncosⁿ⁻¹(x) sin(x) +n− 1 n

Z

cosⁿ⁻²(x) dx;

(iii)

Z 1

(x²+ 1)ⁿ dx= 1 2n − 2

x

(x²+ 1)ⁿ⁻¹ +2n − 3 2n − 2

Z 1

(x²+ 1)ⁿ⁻¹ dx.

Hint: Schrijf 1

(x²+ 1)ⁿ = 1 + x²− x²

(x²+ 1)ⁿ = 1

(x²+ 1)ⁿ⁻¹ − x² (x²+ 1)ⁿ. Oplossing.

(i) R sinⁿ(x) dx =R sinⁿ⁻¹(x) sin(x) dx

= − sinⁿ⁻¹(x) cos(x) +R (n − 1) sinⁿ⁻²(x) cos(x) cos(x) dx

= − sinⁿ⁻¹(x) cos(x)+(n−1)R sinⁿ⁻²(x) dx−(n−1)R sinⁿ⁻²(x) sin²(x) dx

= −n¹ sinⁿ⁻¹(x) cos(x) + ⁿ⁻¹_n R sinⁿ⁻²(x) dx;

(ii) R cosⁿ(x) dx =R cosⁿ⁻¹(x) cos(x) dx

= cosⁿ⁻¹(x) sin(x) +R (n − 1) cosⁿ⁻²(x) sin(x) sin(x) dx

= cosⁿ⁻¹(x) sin(x)+(n−1)R cosⁿ⁻²(x) dx−(n−1)R cosⁿ⁻²(x) cos²(x) dx

= ¹_ncosⁿ⁻¹(x) sin(x) + ⁿ⁻¹_n R cosⁿ⁻²(x) dx;

(iii) R ₁

(x²+1)ⁿ dx=R ₁

(x²+1)ⁿ⁻¹ dx−R x² (x²+1)ⁿ dx.

Er geldt (_(x2 ¹

+1)ⁿ⁻¹)^′ = −(n − 1)_(x^22x₊₁₎ⁿ, dus is R x²

(x²+1)ⁿ dx =R x_(x2^x

+1)ⁿ dx= x_2n−2⁻¹ _(x2 ¹

+1)ⁿ⁻¹ +_2n−2¹ R ₁

(x²+1)ⁿ⁻¹ dx en dus R ₁

(x²+1)ⁿ dx=R ₁

(x²+1)ⁿ⁻¹ dx+_2n−2¹ _(x2+1)^{x n} − _2n−2¹ R ₁

(x²+1)ⁿ⁻¹ dx=

1 2n−2

x

(x²+1)ⁿ +²ⁿ⁻³_2n−2R ₁

(x²+1)ⁿ⁻¹ dx.

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